The radius of the base of a cylinder is decreasing at a rate of $9$ millimeters per hour and the height of the cylinder is increasing at a rate of $2$ millimeters per hour. At a certain instant, the base radius is $8$ millimeters and the height is $3$ millimeters. What is the rate of change of the surface area of the cylinder at that instant (in square millimeters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $155\pi$ (Choice B) B $-310\pi$ (Choice C) C $310\pi$ (Choice D) D $-155\pi$ The surface of a cylinder with base radius $r$ and height $h$ is $2\pi rh+2\pi r^2$.
Explanation: Setting up the math Let... $r(t)$ denote the base radius of the cylinder at time $t$, $h(t)$ denote the height of the cylinder at time $t$, and $S(t)$ denote the surface area of the cylinder at time $t$. We are given that $r'(t)=-9$ and $h'(t)=2$ (notice that $r'$ is negative). We are also given that that $r(t_0)=8$ and $h(t_0)=3$ for a specific time $t_0$. We want to find $S'(t_0)$. Relating the measures The measures relate to each other through the formula for the surface area of a cylinder: $S(t)=2\pi r(t)h(t)+2\pi [r(t)]^2$ We can differentiate both sides to find an expression for $S'(t)$ : $S'(t)=2\pi[r'(t)h(t)+r(t)h'(t)+2r(t)r'(t)]$ Using the information to solve Let's plug ${r'(t_0)}={-9}$, ${h(t_0)}={3}$, ${r(t_0)}={8}$, and $C{h'(t_0)}=C{2}$ into the expression for $S'(t_0)$ : $\begin{aligned} S'(t_0)&=2\pi[{r'(t_0)}{h(t_0)}+{r(t_0)}C{h'(t_0)}+2{r(t_0)}{r'(t_0)}] \\\\ &=2\pi[({-9})({3})+({8})(C{2})+2({8})({-9})] \\\\ &=-310\pi \end{aligned}$ In conclusion, the rate of change of the surface area of the cylinder at that instant is $-310\pi$ square millimeters per hour. Since the rate of change is negative, we know that the surface area is decreasing.